How To Solve A TOUGH Interview Question - Ways To Give 11 Coins To 3 People

How many ways can you divide 11 coins to 3 people? How many ways are there if each person has to get at least 1 coin? This puzzle is an example of a technical interview question. The video demonstrates the solution, which indicates the number of non-negative (and positive) integer solutions when you have R variables adding up to N.
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[@andrewkwasek1214]

If i am applying for some kind of upper management position it is simple. I give them each one and i take the rest.

[@ProfessorSyndicateFranklai]

Or If anyone's applying to any other position, just give them each one and give the rest to the boss.

[@tuhaggis]

*You delegate someone to give each person 1 each and to give you the remainder.

[@matthewfedoseev580]

The Russian government way (I’m Russian, I know my government)

[@help4343]

In Soviet Russia coin distributes YOU!

[@arnavnarula1534]

For those not understanding the video's solution, I think it is better to think of an analogous problem: *1. How many ways can 3 non-negative integers sum to 11?* *2. How many ways can 3 positive integers sum to 11?* For #1: Pretend we have 11 1's like this: 1 1 1 1 1 1 1 1 1 1 1 By putting two dividers between 1's, we see that we can split the number of 1's that each person gets. Ex: 1 1 1 1 (divider) 1 1 1 1 1 (divider) 1 1 --> Translates to 4 + 5 + 2. If we put a divider before the first 1, the first integer = 0 If we put a divider after the last 1, the last integer = 0 In addition, we can have two consecutive dividers; the middle integer = 0 Thus, the question becomes how many ways can we arrange 11 1's and 2 dividers. This is basically the same question: *How many 13-letter unique words can be made with 11 A's and 2 B's?* If you know Combinations: (11 + 2)! / (11! * 2!) = 13! / (11! 2!) = 13 * 12 / 2 = 156 / 2 = *78 ways* If you don't know Combinations, think about it this way: You can arrange the 11 1's and 2 dividers in 13 * 12 * 11... * 3 * 2 * 1 ways. This works because the first letter can go anywhere in the 13 spaces for the word, the second letter will have 12 spaces then to choose from, followed by 11 for the third letter, etc. However, this includes repeats, as swapping two A's still give the same word. Just like two A's can be arranged 2 * 1 ways --> Second A, First A or First A, Second A: 11 A's can be arranged 11 * 10 * 9... * 3 * 2 * 1 ways 2 B's can be arranged 2 * 1 ways Since the above ways are repeats, we need to divide them from 13 * 12 * 11... 3 * 2 * 1 ways (Total Possibilities with repeats) You see that when we divide 13 * 12 * 11 (Total Possibilities including repeats).... by 11 * 10 * 9 (Repeats for A's...), we can call 11 * 10 * 9... = x and 13 * 12 * 11... as 13 * 12 * x. Dividing them, you get (13 * 12 * x) / x = 13 * 12 = 156 ways. Now, we divide again by 2 * 1 (Repeat for B's...) = 2. 156 / 2 = *78 ways* For #2: Again, Pretend we have 11 1's like this: 1 1 1 1 1 1 1 1 1 1 1 Unlike the first question, we can't put a divider before the first 1 (First Integer MUST be positive), we can't put a divider after the last 1 (Last Integer MUST be positive), and we can't have two consecutive dividers (Middle Integer MUST be positive) Then, the only place the dividers can go is between the 1's, and not overlapping. We have 2 dividers and 10 spaces between the 11's. The first divider can go in any of the 10 spaces. The second divider can go in the remaining 9 spaces (Not OVERLAPPING) This gives 10 * 9 = 90 ways. However, the first divider and second divider can switch positions and still count as the same set. EX: 1 1 1 1 (1st divider) 1 1 1 1 1 (2nd divider) 11 & 1 1 1 1 (2nd divider) 1 1 1 1 1 (1st divider) 11 are BOTH STILL 4 + 5 + 2 Therefore, we divide 90 ways / 2 = *45 ways*

[@ryanbeck5522]

In what kind of job interview would you be required to know this? Are you given a written exam? I'm honestly just curious.

[@sussykanyeballs176]

It's to check your quick thinking skills and logic I suppose

[@vjekokolic9057]

I'm guessing they'd have to give you a paper and a pen or a board and a marker, because it's quite impossible to do the maths by head

[@XJWill1]

@@vjekokolic9057 It is certainly possible to do it in your head. I did. I first quickly solved the case for X coins to 2 people in my head. The answer is just X+1 since the first person can get 0 to X coins, and the second person gets what is left. Now, for 3 people, the first person gets X coins and the others get 0 which is 1 way. If the first person gets X-1 then the others distribute the 1 coin which is 2 ways. First X-2 and the others distribute 2 coins which is 3 ways. See the pattern? The answer will be 1+2+...X+1. And I happen to recall the formula for the sum of 1 to N is N*(N+1)/2 but that can be easily derived in your head using Gauss' method (partition into N/2 pairs of N+1 for example N=10 you have (1+10)+(2+9)...(5+6) ). Anyway, the answer to the first question is N=X+1=11+1=12 and N*(N+1)/2 = 12*13/2 = 78. Easy to do in your head. Using the X-3 trick for the second part, we plug N=11--3+1=9 into the formula to get 9*10/2=45 Of course, any decent interviewer will not just want an answer but will want to hear your thinking process. So I would basically explain what I just wrote to the interviewer rather than just sit quietly for a minute and then say 78 and 45.

[@shanoc5902]

Honestly? I don't see the problem as presented actually being an interview question. Now if the question asked how *you* would split the coins in order to pay your employees? Now they're testing something. If you give each employee 3 coins and keep two for yourself you're preventing employee strife but they know that you will act more in your own self interest. Maybe you give one person 3 coins because they did the least work and the other two four coins and intend to keep things quiet. Or maybe you give them all 3 coins and use the last two coins as a bonus. How you approach this problem now can tell the interviewer about what your priorities are, how you'll approach conflict, and stuff like that. Being able to logically figure out every potential mathmatical solution is less important to most jobs than knowing how you would deal with solving the problems in front of you.

[@fex144]

I just wanted to drive a bus and they asked me about poly ... poly-dy-nobials ... pomy-nobilities?! Tough interview.

[@xnick_uy]

I think there's something missing from the explanation of the divider's method: dividers are not supposed to be distinguishable. As presented in the video, an additional constraint should be that the blue divider should appear at the left of the red one. Were they both black, say, this situation would not arise, and the whole reasoning is still valid. It further would allow to explain the division by 2 in 13*12/2 = 78 at 2:59 .

[@benjaminbatt4464]

+x nick you need more upvotes, because this is exactly why it's not simply (n+1)². Without your constraint, it says any divider can be in any spot next to a coin (so 11x left of the coin + 1x right to the last coin), for two dividers this would be 12 positions for divider 'blue' and another 12 positions for divider 'red' (!) for any valid position of 'blue' (!), therfore 12² (-> (11+1)² -> (n+1)²)

[@gabenuss3063]

Yeah, that is indeed a very important detail. Essentially, we're trying to find out how many multisets of length 11 we can get from a set of length 3. This is a "stars and bars" problem. The dividers (bars) and items (stars) should BOTH be identical. For example, *******|**|** means 7 coins go to Person A, 2 coins go to Person B, and 2 coins go to Person C.

[@DariaTimurovich]

the more correct formula for the combination (13 over 2) is: = 13! / ( 2! * (13-2)! ) replace factorials by multiplications: = 1*2*3...*13 / ( (1*2) * (1*2*3...*11) ) divide by (1*2*3...*11) both up and down parts, and (1*2) become 2: = 12*13 / 2 Done.

[@infinitiv525]

(13!/11!)/2=78 (10!/8!)/2=45

[@Dark3nedDragon]

No u, no calculators

[@JohnDoe-rr9mg]

+Resurgence Of The Shadow Dragons you don't need one if you know what a factorial is

[@MrTinnar]

Maaan, to be algebraically corrrect it must be 2!, not just 2...even despite it's the same number.

[@rados123]

That’s why you shouldn’t use calculator while learning math only pen and piece of paper. It makes you think not only input a lot of almost random numbers till u get right answer.

[@indian6755]

factorial came as rescue

[@AleksandrVasilenko93]

Split evenly and the company keeps the remaining? This way you embrace equality and make a profit.

[@IWasHereFirst2]

I think there is a much shorter and better answer to this and it goes as follows (labeling the persons as A,B,C): I count all possible cases how to share at most 11 coins between A and B (in each case the number of coins for C is determined). If person A gets 0 coins, then there are 12 possibilities how many coins person B can get (0,1,2,..,11) . If person A gets 1 coin, then there are 11 possibilities how many coins person B can get (0,1,2,...,10). ... If person A gets 10 coins, then there are 2 possibilities how many coints person B can get (0,1) If person A gets 11 coins, then there is 1 possibility left for person B (0). Thus, there are 12+11+10+...+1= 6 * 13 = 78 possibilities. Analogue for the second case you get If A gets 1 ... B has 9 possibilities If A gets 2 ... B has 8 .. So you get 9 + 8 + ... + 1 = 9*5 = 45

[@brettchr777]

Love the idea of inserting dividers into the mix to make the math faster, but I think you should have explained (at 2:59 and 5:52 marks) where you came up with (13)(12) / 2 ... and ... (10)(9) / 2 ... respectively, and that the choose function (nCr) is calculated as follows: (n!) / [ (r!) (n-r)! ]

[@rhythmpankhaniya3650]

Brett Christoffersen concept of Combination.. yeah it migh be... but i think ans would be (11!/3!)

[@antoniostangaciu2564]

Hi why you cant use the number of functions? 3 at Power 11.

[@imi___]

I got to the same solution with a different reasoning for the first question, so for those of you who don't immediately remember binom numbers or don't like formulas, here it goes: First the observation that "splitting coins between 2 people" is dead simple: For n coins, there are n+1 ways to split them (first gets none, first gets one, first get two... first gets all) With that in mind, lets think about 3 people. The first person can get 0 to 11 coins here. So if the first person gets all 11 coins, there are 0 coins left for person two and three to get. So there are only one way of splitting n=0 coins into two persons: n+1 = 1 way. If the first person gets only 10 coins, there is one coin left for person two and three, that makes another 1+1=2 ways. If the first person gets only 9 coins, there are two coins left, which are 2+1 = 3 ways of splitting there.. ... So if we sum up all the results for when the first person gets the coins, you get the sum from 1 to 12 = 78. (for the second question, I agree with the trick to just give one coin to each one and then come to "sum of 1 to 9 = 45")

[@Fogmeister]

If you are asked this in a technical interview... get up and walk out. It's a shit question that only a shit company would ask. It shows absolutely nothing to do with your ability as a programmer/coder. This sort of whiteboard coding interview question is completely ridiculous.

[@PattyManatty]

+Oliver Foggin Someone's mad. It shows how you approach problems. Generic problem solving is important.

[@Fogmeister]

PattyManatty it's widely regarded that interview questions like this are a bad thing. techcrunch.com/2015/03/21/the-terrible-technical-interview/ Here's a much better alternative that will tell any company a lot more about a developer than terrible whiteboard coding questions.

[@robin888official]

It's a little confusing to distinguish between the two dividers (as blue and red) and than coming up with a formula which doesn't distinguish between objects. I know this "trick" (aka derivation of the hypergeometric distribution) and still was confused. Sounded like the "red" divider could end up left of the blue one.

[@killedragonyt6353]

Imagine that you give 1 coin to blue, then into the red and green you can give 1_9 2_8 3_7 etc., so there are 9 posibilities; If blie get, 2 coins, there are 8 posibilities, with 3 there are 7 posibilities, etc., until the 9, because would be 9_1_1 with 1 posibility of 9; So: N = 1+2+3+4... , all in order, until he reach the number of the high posible way for yhe blue that is toral of coins - 2 = 11 - 2 = 9; So, basicaly: *9+8+7+6+5+4+3+2+1 = 45* xd

[@elizabethnichols9048]

441 369 I was way off

[@ItaiSitnik]

And if you want only unique solutions that are not repetitive? I mean, if I say that: person1: 8 person2: 2 preson3: 1 person1: 2 person2: 1 preson3: 8 are the same combination

[@breathless792]

+Itai Sitnik theres 16 possibilties and 10 when each person has at least one

[@ItaiSitnik]

+breathless792 how do you calculate that parametrically?

[@MindYourDecisions]

+Itai Sitnik I believe this would be the number of ways to partition N into exactly k parts: en.wikipedia.org/wiki/Partition_(number_theory)#Restricted_part_size_or_number_of_parts

[@ItaiSitnik]

+breathless792 how did you managed to create a sheet like this?

[@breathless792]

basically I made a table of all possibilties of 3 numbers between 0-11 added together (I had a short cut to writing it but its a bit complicated) and then used the autosum function then sorted them in order of the sum so I could isolate those whose sum was 11 them used another slightly complicated method to remove the duplicates.

[@visantibanez]

I love your videos!! MindYourDecisions, Numberphile and 3Blue1Brown are my favorite math KZbin channels. Good job 😄

[@targetiitbcse1761]

Watch Michael penn

[@GothicKin]

For no apparent reason I could never quite work out the stars and bars method even though I could naturally come up with more complicated maths. That is obviously until this video, as always a great explaination!

[@joshc8599]

i like mine the best, each gets one and i get eight. they dont need to know about those other eight

[@bethcary1003]

So I miss-understood the problem. When you said WSU’s to divide, I thought that 0-0-11 (etc) would only count once because I was counting the ways to divide 11 into 3 groups.

[@gblargg]

Same, I took the people being generic, and came up with 15 combos for part 1, 9 for part 2. It was simple enough to just enumerate the possibilities by hand.

[@jozsika969]

Such vaguely defined and utterly useless problems. Gosh!

[@itsavinashtiwari]

its not tricky its just a permutation combination problem..basic knowledge of p n c is suffice..

[@stonythewoke9921]

yeah its not a permutation tho, its a combination

[@purplecoathanger]

I saw this is as an arithmetic series: (n / 2) (2a + (n - 1)d) for part 1: n = 12, a = 1, d = 1 for part 2, n = 9, a =1, d = 1 (could've been written more compactly as (n^2 + n) / 2 in this case)

[@rajeshgupta1055]

Good stuff, but the admin need to know that we are not kindergarten children that he has to explain us each and everything like 1+1=2. just jump to the point and skip obvious parts.

[@robertomasymas]

How did you get to 13 instead of twelve? I mean I can see that you choose slots to place either coins or dividers in, but at first you were looking at the spaces between the slots - you would also need spaces on the outer ends to allow for 0 coins to one person. So MY solution was 12c2 instead of 13c2. The difference between the two approaches is that in the video the solution creates slots for either coins or dividers, while the solution I gravitated towards looks at the spaces between and around 11 coins. Somehow you have to know that you can't technically put two dividers in one space. Im not sure how that is obvious.

[@bene2451]

the first is equivalent to 1+2+3...+(n+1)

[@78anurag]

Dude ez just use the combinational formula GG

[@fedeh777]

shouldnt the calculation be 11!/((11-3)!*3!)??

[@karlmarxstolemybike3382]

What kind of interview asks a question like this??

[@d0u6la5m]

Give all the coins to the red man because you know blue man and green man dont exist. OR say they do exist and you charged each man 10 coins each to solve this puzzle so their total spent would be 30 coins. But then you change your mind and decide to charge them 25 coins instead of 30 so you try to give them back 5 coins but cant do it evenly. Therefore you give them back 1 coin each and pocket the other 2. Each man has now spent a total of 9 coins each. 9x3=27 plus the two in your pocket is 29. Now there is a missing coin.

[@TheKivifreak]

Great trick question. They gave you 3x9 coins. 25 + 2 in your pockets. The remaining three you returned, so no coin is missing. 25 + 2 + 3 = 30. It took me a little while to see through your wording.

[@Dark3nedDragon]

No, you gave each man back a coin, so they have three total, and you put two in your pocket, leaving 25 out in the open, 2 in your pocket, and 3 returned back to them, totaling 30. It has nothing to do with a trick question, just bad math at the end. 30 - 5 = 25. You pocketed two, they each paid 10, so 10 x 3 = 30. You returned one coin to each of them, so they each paid 9, but 2 of the 5 you kept, leaving the total they gave you at 27, not 28. Adding back in the three missing results in 30. It wasn't the wording that could confuse someone, rather the bad math given instead.

[@TheKivifreak]

Kudos on answering with the same answer and twice the word count. Sorry for teasing :^)

[@aliali-hk3dm]

I ate the missing chocolate coin

[@ProfessorSyndicateFranklai]

Actually, one of the premises is wrong: 25/3 is not 9. It's 8 1/3, plus you gave them one coin each, but you pocketed two. In reality, they spent 28 coins total, or 9 1/3, plus the 2/3 coin fee. Your logic was infallible until the premises was proven wrong.

[@durgaprasad807]

I used binomial theorem to solve this

[@ghffrsfygdhfjkjiysdz]

If they really did ask questions like that on job interviews, we'd have 100% unemployment.

[@ProfessorSyndicateFranklai]

Actually, he solved it, so it'd really be more like (p-1)/p let p = the population of the selected group.

[@vaibhavnandha7195]

way underrated

[@AcheiversOfficial]

looks like you have explained a simple question in a more complicated manner..:-)

[@BravoNineThreeTwo]

A lot of overly complicated explanations. A more simple solution. Q1: How many ways can you split 11 coins between 2 people? Q2: If you gave one coin to a third person, how many ways could you split the remainder between the other two? Q3: What’s 12+11+10+9…

[@titotitoburg6298]

i would like to get the equations better explained. Why can't we do 11 choose 3? what's the difference? I've never heard of these binomial expressions, i want the algebraic explanations please.

[@alansands256]

Because 3 dividers creates 4 divisions. But there are still just 3 people.

[@0cheeseburga]

wait, was I supposed to assume 11blue 0red 0green is different than 0blue 0red 11green? to me thats not a different way to divide the coins but rather a way to distribute them

[@michellehunt368]

That's what I thought. Unique combinations is different. I tried to solve that problem (ie if 0 0 11 was a result 11 0 0 was not) how do we solve that with an equation?

[@entropiCCycles]

Well, I overestimated, then. I tried imagining the parameter space as a 3-simplex instead of the 2-simplex it actually is.

[@elieldushime6324]

What if we change the question a little bit. First let's keep : "each person has to get at least 1 coin." But then add : " each person has to get at most 7 coins." ?????????????

[@elieldushime6324]

it becomes more difficult,but I'm sure some will find the solution.good luck with that

[@yurenchu]

@RadkeMaiden, Wow, cool! Thanks for posting that, I didn't know that you could solve it like this. Very clever. I guess the method also applies when each person has his own restrictions on the number of coins he gets: just include or exclude the polynomial terms with the appropriate exponents.

[@theonesparrowhawk]

This isn't a general solution. For the problem in the video, I listed all the cases which weren't a permutation of other cases (e.g. 4,5,2 is just 2,4,5 just switched around). Then I found out how many ways each of the distinct cases could be distributed (if two numbers are identical then there are only three permutations, while if all three numbers are different then there are 6 permutations. So I just need to remove the cases from my list where someone has more than 7 coins. The number of cases that follow your new rule, including permutations, is 36 - removing 9,1,1 (3 permutations) and 8,2,1 (6 permutations) from the previously stated 45 cases.

[@brucefulton]

I appreciate what you are trying to do, but you make a lot of misstatements that you then correct, which makes it very hard to follow. It would be much easier to follow if you would edit those out.

[@TrebleWing]

Just cause math is hard, it doesn't mean you should down vote the video. I don't get a single thing he's saying but I'm not about to dislike the video just cause I'M an idiot.

[@MichaelKondrashin]

This can be solved "straight ahead" without using "choose" stuff. N = 11. First answer is sum for i from 0 to N of sum for j from 0 to N-i of 1. i is number of first person coins, j is second person coins and third one gets the rest of coins, whichever number of coins remains. This sum is easy to calculate if you know how to calculate arithmetic progression. Second answer is similar: Sum for i from 1 to N-2 of sum for j from 1 to N-i-1 of 1. In this case there need no need for heuristics for blue and red divider

[@shashankkothari8066]

This can create confusion for a lot of people, We need two positions out of 12 slots to divide 11 coins between 3 people. And because the slots are identical, No. of ways=¹²C² But because we also want 12 such cases when both slots picked are exactly same positions. Therefore, Total no. of ways=¹²C² + 12. General formula for n no. of coins, General formula=(n+1)C² + (n+1) =(n+1)C² +(n+1)C¹ =(n+2)C²

[@cga490]

Hey, quick question, don't you need to limit somehow the nCk (or nPk I always confuse them) so the red divider is always on the right of the blue one?

[@MindYourDecisions]

+Carlos García Ávila I can see how that's confusing, perhaps I should have done the video a bit differently. Basically the 2 dividers are identical--so I could have colored them black. So imagine placing 2 identical dividers, and then coloring the one the the right as red and the one on the left as blue.

[@mightyassad]

+MindYourDecisions For the first answer, shouldn't it be like 12 * 12 / 2 = 72, coz there are 12 places where each divider can be placed ?

[@cbernier3]

+Assad Mangou No, there are 12 places for the first one. After that, there are only 11 places left for the second.

[@mightyassad]

+Zabjex Thanks for your explanation... That's true, i didn't presume about both being on the same slot :)

[@ywlin19891]

There are 12 cases where the dividers are in the same gap, so you don't divide them by 2. Thus 12 * 11 / 2 + 12 = 78

[@jdfoehl]

A simpler way: 6 distinct combos with 0... each combo has 6 variations except 0 0 11 which has 3 because a digit is repeated. 5 distinct combos with 1 and no 0 ... 3 with 6 variations and 2 with 3 3 distinct combos with 2 no 1 or 0... 2 with 6 variations and 1 with 3 2 distinct combos with 3 no 0, 1 or 2... with 3 variations. Add up all variations ... 6 variants :10 = 60 + 3 variants: 6 = 18 = 78

[@sammy45654565]

i have a more intuitive method that doesn't involve binomial coefficients. if you imagine all the scenarios where person 3 has one coin, you can start with person 1 having one coin and person 2 having nine. then move person 1 up to two coins with person two moving down to 8. if you move across the line until there are two coins remaining (person 1 having 9 coins with person 2 and 3 having one coin each), you end up with 9 different scenarios where person 3 has one coin. (in other words, start with the pointer finger of your left hand on coin one, representing how many coins belong to person 1 and move your finger across until there are two coins left. the number of different finger placements is the number of different scenarios with person 3 having one coin) in order to determine how many scenarios there are with person 3 having two coins, you repeat the process with one less coin (with that coin automatically going to person 3) and end up with a number of scenarios that is one less than the previous situation. so there are 8 different scenarios where person 3 has two coins. (in other words, cover the right most coin with your right hand pointer finger, and repeat the process you underwent with your left pointer finger until there are two coins left) repeat this two step two-finger process all the way through until person 1 has one coin, person 2 has one coin, and person 3 has 9 coins. you end up with a sequence of 9+8+7+6+5+4+3+2+1 = 45 different scenarios. the answer of 78 happens because allowing the "division" of coins to zero for one, two, or three people means there are 3 extra phases to the sequence. (12+11+10)+9+8+7+6+5+4+3+2+1 = 78

[@poznyakpoznyak]

A)First guy can get from 0 to 11 coins. Then x=11-a coins are left, which we split between two other guys. There are x+1=12-a ways to do it: x:0, (x-1):1, (x-2):2, ..., 0:x. So the number of ways is [a=0,11]sum[12-a]=[b=1,12]sum[b]= 12*13/2 equals 78. B)If each person has to get at least 1 coin, we just give 1 coin each right away, and then we can split 8 remaining coins the way we did it in A. [a=0,8]sum[9-a]=[b=1,9]sum[b]= 9*10/2 equals 45.

[@firatcubukcu6984]

You can solve it without using any equation by only writing down the possible distributions. It only took me 3-4 minutes to solve it. The numbers in brackets represent how many times this order can be distributed. If a number is a duplicate there are 3 ways, if every number is unique there are 6. First, we write down the possible distributions involving zeros 11 0 0 (3) 10 1 0 (6) 9 2 0 (6) 8 3 0 (6) 7 4 0 (6) 6 5 0 (6) Then we write down the possible distributions with duplicates: 1 1 9 (3) 2 2 7 (3) 3 3 5 (3) 4 4 3 (3) 5 5 1 (3) The possible distributions where every number is unique: 1 2 8 (6) 1 3 7 (6) 1 4 6 (6) 2 3 6 (6) 2 4 5 (6) So when we sum up all of the numbers in brackets, it gives us the answer to the first question: 78. Excluding the ones that involve a zero, we get to the answer to the second question: 45.

[@david21686]

Edit: Nevermind; I did the math correctly a second time and got the same answers as you. I'm leaving my original comment up because it shows a second method of solving the problem. ------ BULLSHIT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! The way you set the problem up, the red divider could have been on the left side of the blue divider. Even if you colored the dividers black, you would end up double-counting all the ways that the black dividers could be divided up. Or maybe the issue isn't with double-counting, maybe it's with something else. But I'm pretty sure you're wrong. Here's how I solved the problem: 1. There are N+1 ways to divide N balls between two people (quick proof: The first person can get 0, 1, 2, 3, ... , N-1, or N balls. The second person gets all the rest. It's basically counting all the balls; imagine the second person giving the first person all the balls one at a time so they can count the balls together. During that counting process, they will eventually see themselves in every scenario where the balls are split up in every single possible way). 2. Now add a third person. 3. If the third person gets all 11 balls, 0 balls remain, and there's exactly 0+1 way to divide the remaining balls between the remaining 2 people 4. If the third person gets 10 balls, 1 ball remains, and there are exactly 1+1 ways to divide the remaining balls between the remaining 2 people. 5. If the third person gets 9 balls, 2 balls remain, and there are exactly 2+1 ways to divide the remaining balls between the remaining 2 people. 6. Continue that pattern, then use the fact that 1+2+3+...+N=N * (N+1)/2, and do the math in your head. ------- Now the second problem. 1. There are exactly N ways to divide N balls between two people so that everyone gets at least one. 2. Now add a third person. 3. That third person needs at least 1 ball. 4. If that third person gets more than 9 balls, then somebody gets 0 balls. 5. Etc. etc. I did the math in my head and got 66 combinations for the first problem and 45 combinations for the second problem. And yet here you are with your fancy schmancy video, trying to mess with my head and make me think I'm getting Alzheimers. Are you happy????

[@MWSin1]

I did it via a sigma rather than a choose. Perhaps more brute force, but once you see the pattern, easy to extrapolate for 3 people. If blue gets 0 coins, there are 12 combinations of what red and green could get (0-11, 1-10, 2-9... 11-0) If blue gets 1 coin, there are 11 combinations of what red and green could get (0-10, 1-9, 2-8... 10-0) If blue gets 2 coins, there are 10 combinations of what red and green could get (0-9, 1-8, 2-7... 9-0) Easy to spot the pattern: If blue gets k coins, there are 12-k combinations of what red and green could get. From there, it's just a simple matter of sigma(12-k) from 0 to 11. In other words, 12*(1+12)/2 = 78. If you were trying to come up with a general solution for any number of people, it would quickly become unwieldy, though.

[@Milkymalk]

Much easier and doable without math beyond fourth grade if you split this problem into stages: How many possibilities are there for how many coins Blue gets? 12 (0 to 11). For each of these possibilities, counted as n, add 12-n ("How many different numbers of coins can Red get?" = "number of coins left +1") to a sum for 12+11+10+9+8+7+6+5+4+3+2+1 = 78. Explanation: If Blue gets 0 coins, then 11 coins MUST be split between Red and Green. As there are 12 possibilities for how many coins Red gets (all of which dictate how many coins green gets), this adds 12 possibilities to the total: 0-0-11, 0-1-10 and so on until 0-11-0. If Blue gets 1 coin, then 10 coins MUST be split between Red and Green. As there are 11 possibilities for how many coins Red gets (all of which dictate how many coins green gets), this adds 11 possibilities to the total: 1-0-10, 1-1-9 and so on until 1-10-0. until: If Blue gets 11 coins, then there are no more coins left to be split between Red and Green. This adds 1 possibility to the total: 11-0-0 So the solution is: sum(1..(number of coins+1))

[@MateoJH]

I got this answer in a much simpler and faster way somehow just by adding 12 possible outcomes for each person if you include 0 coins. 12+11+10+9+8+7+6+5+4+3+2+1 = 78 If each person must have 1 coin, it's only 9 possible outcomes for each person. 9+8+7+6+5+4+3+2+1 = 48 I think this was somehow explained at 3:43 and 6:33 I guess I just did a whole heck of a lot of work in my head subconsciously and/or got really freaking lucky. lol

[@davidlee4314]

I thought if this as if figuring out how many numbers between 0 and 999 have their digits add up to 11 or 8. It follows the trend of the sum of all positive integers less than and equal to the sum (11 or 8 in these examples) plus 1. So for 1 coin it’s 3 (1+2), for 2 coins it’s 6 (1 + 2 +3) for 4 coins it’s 10 (1+2+3+4) and so on.

[@LillianWinterAnimations]

(n - 1) * (n - 2) * (n - 3) * ... (n - ( r - 1 )) = N Here's my answer for the number of solutions for the end with the extra condition applied. n = number of coins r = number of people N = number of possible configurations (INCLUDING symmetrical answers (1, 2, 3 and 2, 3, 1 for instance))

[@theMosen]

Video paused: So are the people allowed to get zero coins? I hate when test questions are not unambiguous! Anyway, I'm gonna say no, cuz that isn't really "splitting". So I'm guessing the answer is 9+8+7+6+5+4+3+2+1 = 45. Add 33 if zeros are allowed, but subtract 3 if a combination like 11 0 0 isn't allowed, because that really isn't "splitting". Edit: Ok, I guess I paused a little too early.

[@olbhd]

The solutions given seem unnecessarily complicated. Each of the two dividers can be in one of 12 locations. So there are 144 possibilities. Twelve of those are unique (where both dividers are in the same location). That leaves 132 other possibilities, which occur in pairs (eg. first divider in location A, second in B, and vice-versa are the same), so that's 132/2 = 66 unique ways. 66+ 12 = 78. I don't see the justification - or the need - in the first method for using extra coin location for the dividers - especially since the graphic which introduces the dividers shows them (correctly) drawn between the coins!

[@edm_ortiz]

It can be more easy; Just 12+11+10+9+8+7+6+5+4+3+2+1, since there are 12 spots where the one divider may go. The other one: 9+8+7+6+5+4+3+2+1since there are 9 spots where the dividers may go (minus 3 because they can not go to the sides and not be on the same spot)

[@oksimoron5246]

You go with the lowest value: 1,1,9 (because there are two same number 1 and 1 there are 3 combinations for that, 1,2,8 (6 combinations) 1,3,7 (6 comb) 1,4,6 (6comb) 1,5,5 (3comb) for 1 in it there are 24. Now with 2,2,7 (3) 2,3,6 (6) 2,4,5 (6)== 15 3,3,5 (3) 3,4,4 (3) ==6 24+15+6=45

[@chriskeck2969]

simple. the first person could be handed anywhere between 1 and 9 coins (leaving at least 1 for each of the two others). If you give the first person one coin, there will be 10 coins for the other two to divide. There are 9 ways you could divide those 10 coins amongst the two people. 1 to the second, 9 to the third is the first option, followed by 2 to the second and 8 to the third. You can do this until you reach 9 for the second and 1 for the third. That means if you give the very first person one coin, you have 9 ways to distribute the other 10 coins to the last two people. Now instead, give the first person 2 coins. Divide the 9 coins between the other two, and you'll find you have 8 ways to do so. Then, if you notice, when giving the very first person 3 coins, the other two have 7 ways to split the remaining 8 coins. Repeat until you cannot repeat anymore. it's basically 9+8+7+6+5+4+3+2+1=45. aka [(9^2)+9]/2

[@jumpman8282]

Giving this video as an answer to a job interviewer would most probably be _disastrous_ to my likeliness of getting the job, since I don't believe that the purpose of the question is to evaluate my combinatorial skills, but rather to find out about my thoughts regarding 'fair share' and also (to some degree) my ability to think 'outside the box'. The "correct" question would be: "How do you divide 11 coins _equally_ among three persons?" The mathematical answer to this question would be to give each person 3 and 2/3 coins, but in real life that's not very applicable, since any fraction of a coin is worthless anyway, and therefor the "best" answer would be to give them three coins each. This, inevitably, leads to the _follow-up question_: "What do you do with the remaining two coins?" The way I see it, I have four (reasonable) options: 1) Keep the two coins for myself. 2) Give the two coins to one of the three persons. 3) Give an additional coin to two of the three persons. 4) Give all three persons an additional coin (given that I myself have a coin to spare). I don't think that the answer itself is very important to the interviewer, but rather my reasoning towards one of these four options.

[@jlegourd]

The question doesn't exclude 0 as a valid share of coin (no giving money is a probable scenario). Similarly it doesn't request to consider people identical (people being identical is not a probably scenario). Hence, for any of the 6 order of people (3*2), there are 66 possible splits (11+10+9...+1). The solution is 6*66=396.

[@srinjenadas3486]

I did the second part in a different way. No. of ways in which each get at least 1 coin = 78 - no. of ways in which 1 or 2 people get 0 coins. So , first if we consider the first person gets 0 coins , then the 11 coins would be distributed over 2 remaining people which is 11C1 i.e. 11 ways. But , the same thing can be done with the second or third people getting 0 coins. So, the total no of ways =3*11= 33 ways. So, the required no. of ways =78-33= 45 ways. :)

[@jesusmind1611]

Pause the video give the problem a try..okay Let 11 coins be split 3 ways is there an equation this community college drop out could develop to represent the problem...let n = the number of combinations and on my paper in long and this just got too big. I am fuking cluless. If A gets 0 there are 11 possible combos and if A gets 1 there are (11-1) more and if A gets 2 there are 11 more until A gets 11 then there is just one more combo. Then do the same for B minus any over lap and again for C minus any overlap which there should be some equation for this I give I am a stupid man Somewhere around three hundred so what?

[@loqiloqi]

The number of possible allocations of C coins among P purses is the number of possible allocations of C-1 coins among P purses plus the number of possible allocations of C coins among P-1 purses. If you make a table of those numbers, you get Pascal's Triangle! I'm hired.

[@jonathanrudolph5045]

For 11 coins you went to a shit ton of trouble to calculate the answer. [As a side note, the question's wording is ambiguous as to whether the solution is for three people without regard to unique identities.] For the first question, all you need to do is determine how many ways can you split up 11 coins for two people because the first will have 0 for every one of those combinations. Figuring out the splits takes, what, 6 seconds? Six combos with the zero at the beginning for the third person, then another two bc of needing to shift the zero right one space two times. Total of 8.

[@johnlally1862]

I was gonna count them out at first and probably give up. Then I saw a pattern and created a mathematical formula that gave me the answers 165 and 195. I think the mathematical formulas I created were already created for a different thing, but they obviously didn't work. What's sadder than a kid that's doing math for fun? Well I'll probably be pretty damn happy about it in the future when I got money and women for having a math job, because Mathematicians make so much money. Plus Mathematicians are great with money.

[@Paxmax]

OOaaahaha, there are so many missing optional sterotypical stick figure side stories to this problem... like: who's making the money handed out here? Why aren't they visible? ..but the answer is... .. .. .. .. 78

[@Dukain01]

I don't quite understand the purpose of the "extra statement". The original question is "how many ways can you split 11 identical coins to 3 people?" So, the extra statement is already implied, because for any instance where 1 of the 3 people do NOT receive a coin, the question isn't satisfied. You just split 11 coins between 2 people, not 3. Or maybe I'm just going crazy.

[@sachinjagarwar]

each of them should get at least one coin ok.then only this will division(fair).or u should mention it before

[@okaro6595]

First you can ignore the green one. He gets what is remaining after the two. If one gives 11 to blue, there is only one way. If one gives 10 there is two (0 or 1 to the red one)... if one gives zero to the blue there are 12 ways (0 to 11 for the red). One gets 1+2+-..+12 = 12*13 / 2 = 78. For the later 9 * 10 / 2 =45.

[@pieterluyten3559]

for the second problem you coulddo this: number of ways to divide coins-number of ways where someone gets 0 coins. if you start by saying 1 person gets no coin you only have to divide the coins with the remaining 2 persons wich gives 11 options. multiply by 3 because thats only the number of options for that 1 person wich gives 33. 78-33=45.

[@amberwilliams6746]

The problem I have with this video (and others by you) is that it is not clear to me that you are asking for all the unique possibilities. What I mean is that I did not realize you counted (for example) Blue getting all 11 and Red getting all 11 as separate ways to divide them, so when I solved it I did not take this into account and thus got a much smaller number. I think you should make that more clear in your videos.

[@Goofyoofydoofy]

What about this riddle Person 1 has any number or coins, Person 2 always has double the amount then person 1, Person 3 always has half of what person 1 and 2 have together (he can get a half coin), How many coins each person have if there is 36 coins?

[@tassay]

your way is ok but i think its more easy if you just ignore the third person. Just say there are only two persons and not all coins must be shared its the same case as if the remaining coins would go to the third. But now you see instantly that its sum of 12 because we have 11 cases. Person A gets 0 ,than there are 12 ways for b( 0 to 11); A gets 1 coin so there are 10 ways for b ( 0to 10) and so on you instantly see , that there are sum of 12 ways to share the coins among the two persons. might be my personal opinion but i found it really easy to see after using that trick.

[@Mastroid-je1kt]

Shouldn't the answer be 78-6 for the six cases were the second to people get the same number of coins twice to make the total number of options, like if the first guy gets 9 then the last options are 2-0, 1-1, 0-2 but with the method shown wouldn't 1-1 be counted twice when it should only be counted once

[@SPUDIKIN]

not that this video wasn't interesting. But I think for interview proposes this is more of a philosophical question than mathematical. because if you HAVE to give all the coins out there will be someone who gets one less. how do you decide who gets left out. common answer would be the least efficient employee. But then the questioning goes on organically with "well how do you decide who's the least efficient?" then so on and so forth.

[@lucaro]

Shouldn't it be (N+r-1)^(r-1) instead of (N+r-1) choose (r-1) ? You need to allow to choose the same spot for both "dividers" as you did in 2:08. Your method would in your (first) example not allow for one person to get all coins. Or am I missing something?

[@DarkKyugara]

QUESTION 01: Could be all those combinations between: (0B, 0R, 11G); (0B, 11R, 0G); (11B, 0R, 0G) and (4B, 4R, 3G); (4B, 3R, 4G); (3B, 4R, 4G) QUESTION 02: Could be all those combinations between: (1B, 1R, 9G); (1B, 9R, 1G); (9B, 1R, 1G) and (4B, 4R, 3G); (4B, 3R, 4G); (3B, 4R, 4G) To both questions, there's only options between those possibilities, there's no way you split 11 coins between 3 people without hit one of those possibilities above.

[@LinkinPark4Lifeify]

I cant understand this. I pressed 11C3 on the calculator like i normally would have done for this type of math problems. But why is it actually 13C3. I mean i have done this million of times back in when i was studying high school.

[@smileyball]

I'm not a fan of the way you presented the solution to second problem (the "at least 1 coin per person" problem). Rather than pre-assigning coins to all members, it's more elegant to consider it as choosing how to put (r - 1) dividers in between the slots for the N coins. The math is the same, but the interpretation is cleaner.

[@harmony6643]

So basically you have 8 coins to give to three people and in the end everyone gets another one. And then we just use (10 2) which equals 45 and that's the number of ways. [WATCHES VIDEO] Exactly.

[@this_mfr]

Even though I passed college Stat with an A, I hated it. I had to carry around a flash card with me all of the time with all of the formulas and definitions on it because statistics and my brain just don't click. I learned the formulas, but I never understood the "whys" of any of it.

[@AnssiRai]

Well, the ideal answer the interviewer probably was after, would have been that the test person takes one extra coin out of his pocket and gives each person four coins, instead of starting to make complicated mathematical calculations that will not add up fairly anyway :D

[@armoredwarfaretutorials7312]

another way do divide the coins is to give each of the three persons a sword and shield. Then tell them to have a gladiator style death match for their share of the coins. as this is obviously a job that had the oddest tasks I have ever heard of a gladiator death match would probably be welcomed at the office. AND it only cost 11 coins, what a bargain!

[@satyasubrahmanyam9496]

okay, in first method if i place the blue divider to the extreme left or red divider to the extreme right, the chances of getting coins for blue & red persons is zero, so we have to leave the extreme positions go satisfy the second part of the question! so the answer would be 11c2. correct me if i am wrong!

[@eduardoteixeira869]

I did it differently and I think is more intuitive solution without using combination. First give to the left person 0 coins, now we have 11 coins to distribute between two persons, this is the same to figure out how many integers sum give me 11, 0+11, 1+10, 2+9..............10+1, 11+0, we can see we have 12 different ways to do that. Now lets assume first person has 1 coin, these let us with 10 coins to distribute between the other two, again we can think as a sum of integers giving 10, 0+10, 1+9, 2+8.................9+1, 10+0, and we see 11 ways to get that. We can further extend this logic and realize that for 0 coins in the first person we have 12 ways to distribute the remaining coins to the other two persons, for 1 coin to the first person we have 11 ways to distribute the other coins, for 2 coins to first person we have 10 and so on we have the following sequence or table, 0->12, 1->11, 2->10, 3->9, ........9->3, 10->2, 11->1. So to find the total number of ways we need to add 12+11+10+9+8+......+3+2+1. This is a PA(Arithmetic Progression) whit the sum (12+1).12/2 where we get 78. For the second part we simply remove 3 coins for the 11 getting 8 and repeat the same thinking and we get 9+8+7+6+........+3+2+1. (9+1).9/2 and we get 45. Thanks

[@neobullseye1]

What if you move the blue divider to the far right and the red one to the far left? Then Blue Guy gets all the coins since theyé all to the left of the blue divider, Green Guy gets all the ones since they're to the right of the red divider and Red Guy gets all the coins because they're in between the two dividers. Of course they then get into a massive argument and fight each other to the death, a fight we will tape and put on the internet for fun and profit. The survivor of the fight gets all the coins, by the way.

[@toadstar1004]

I just gave blue 0 coins meaning there were 12 ways to distribute the coins between red and green. Then when Blue had 1 coin, there were 11 ways. So I added 12+11+10...+2+1 to get 78. For the extra condition, I subtracted 12 from 78, because that was when we gave blue 0 coins. Then I subtracted 1 because there's only 1 way to distribute the "remaining" coins when we give blue 11 coins (0 to red and green). Lastly, I subtracted 20, because 10 remaining ways to give blue coins * 2 remaining coin distributions (0 to red or 0 to green) = 20. 78-20-1-12= 45.

[@silviamayo2993]

After the first part of the problem, I thought the answer to the second part would be 72, since there are 6 ways in the first part where someone gets 0 coins. Have I missed something? Why does that not check out?

[@liberty1425]

I paused and figured it out my own way then continued the video and just got so lost and confused with all the numbers and math I got it correct but still felt like an idiot because I did it differently without math and letters and division but yaaay I got it right 78 and 45

[@NehemiahKiv]

My method I thought was pretty good as well, granted not nearly as fast, i didn't figure out the slots. But for the 2nd portion of the problem. I just wrote out, 1,2,8. 1,3,7. etc. then the same for 2 and 3. For all the non-repeating 3 digits, i know there are 6 unique combinations. Half of that if there is a repeating number. So by starting with 1, you get 24 combinations. 2, gets you 15 unique combinations. 3, gets you a total of 6 combinations. In total I got 45 and it was quicker than watching his explanation, not sure if i would get the job though

[@dreadpiratetjoris]

By making them blue, red and green, each person is unique, so you should count double (triple) results. For instance a situation where ‘a’ person gets all 11 coins should be counted 3 times, once for each color. Is that considered in the 78 solution?

[@Anduril919]

I think your first solution is wrong by 1. To divide 11 coins among 3 people assumes that at least 1 of them gets something. So placing the divider on the 1st and 2nd position is excluded.

[@bananasunshine]

And here you specify that the coins are identical, when it doesn't even matter? While in the socks question where it actually matters if they're identical, you didn't specify and just assumed.

[@mander40101]

You don't usually stutter during your explanations, so it's pretty obvious this confused you. Not surprising, since I understand none of this.

[@Rekko82]

Jay-Z gets 6 coins, Beyonce gets 5 coins, I get 0 coins Beyomce gets 6 coins, Jay-Z gets 5 coins, I get 0 coins This is the basics of equality - so there are 2 answers in reality.

[@TheRevScare]

Pretty interesting. If both the coins AND the people are identical (such that it doesn't matter which person gets which coin), then the answer to the first part is 16, and 10 when each person has at least one coin.

[@Fernando_Cabanillas]

there is a pretty basic error in this calculation, according to your math, the blue and red separators could swich sides giving repetitive answers (or negative answers depending on how you operate it) since you have no condition to keep blue to the left and red to the right

[@ultratrololol]

How would this work if you wanted to split 3 coins to 11 people? Or generally if the people are more than the coins? (Of course some people would get 0 coins)

[@GroovingPict]

13! / (11! * 2!) and 10! / (8! * 2!)

[@MrDOOlevrai]

No need to shout, maaaaan

[@AngirasDarbha]

Thats not him shouting, Jackass It's a mathematical function known as factorial, which means the product of all whole numbers lesser than and equal to x

[@ThatOneKid755]

+Angiras Darbha You don't understand humor.

[@MrDOOlevrai]

Angiras Darbha Equal to kiss? How is that math?

[@psychwolf7590]

Mr.DOO cringe

[@farberbrodsky]

Nice! This is exactly what my lecturer taught us in our first combinatorics class :)